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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Example 30 (Method 1) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane.The equation of a line passing through two points with position vectors ๐‘Ž โƒ— & ๐‘ โƒ— is ๐’“ โƒ— = ๐’‚ โƒ— + ๐œ† (๐’ƒ โƒ— โˆ’ ๐’‚ โƒ—) Given the line passes through the points (๐‘ โƒ— โˆ’ ๐‘Ž โƒ—) = (5๐‘– ฬ‚ + 1๐‘— ฬ‚ + 6๐‘˜ ฬ‚) โˆ’ (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 1๐‘˜ ฬ‚) = 2๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚ A (3, 4, 1) ๐‘Ž โƒ— = 3๐‘– ฬ‚ + 4๐‘— ฬ‚ + ๐‘˜ ฬ‚ B (5, 1, 6) ๐‘ โƒ— = 5๐‘– ฬ‚ + 1๐‘— ฬ‚ + 6๐‘˜ ฬ‚ โˆด ๐‘Ÿ โƒ— = (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 1๐‘˜ ฬ‚) + ๐œ† (2๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚) Let the coordinates of the point where the line crosses the XY plane be (x, y, 0). So, ๐‘Ÿ โƒ— = x๐‘– ฬ‚ + y๐‘— ฬ‚ + 0๐‘˜ ฬ‚ Since point crosses the plane, it will satisfy its equation Putting (2) in (1) x๐‘– ฬ‚ + y๐‘— ฬ‚ + 0๐‘˜ ฬ‚ = 3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 1๐‘˜ ฬ‚ + 2๐œ†๐‘– ฬ‚ โˆ’ 3๐œ†๐‘— ฬ‚ + 5๐œ†๐‘˜ ฬ‚ x๐‘– ฬ‚ + y๐‘— ฬ‚ + 0๐‘˜ ฬ‚ = (3 + 2๐œ†)๐‘– ฬ‚ + (4 โˆ’ 3๐œ†)๐‘— ฬ‚ + (1 + 5๐œ†)๐‘˜ ฬ‚ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 5๐œ† โˆด ๐œ† = (โˆ’๐Ÿ)/๐Ÿ“ So, x = 3 + ๐œ† = 3 + 2 ร— (โˆ’1)/5 = 3 โˆ’ 2/5 = 13/5 & y = 4 โˆ’ 3๐œ† = 4 โˆ’ 3 ร— (โˆ’1)/5 = 4 + 3/5 = 23/5 Therefore, the required coordinates are (๐Ÿ๐Ÿ‘/๐Ÿ“,๐Ÿ๐Ÿ‘/๐Ÿ“,๐ŸŽ) Example 30 (Method 2) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane.The equation of a line passing through two points A(๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) and B(๐‘ฅ_2, ๐‘ฆ_2, ๐‘ง_2) is (๐’™ โˆ’ ๐’™_๐Ÿ)/(๐’™_๐Ÿ โˆ’ ๐’™_๐Ÿ ) = (๐’š โˆ’ ๐’š_๐Ÿ)/(๐’š_๐Ÿ โˆ’ ๐’š_๐Ÿ ) = (๐’› โˆ’ ๐’›_๐Ÿ)/(๐’›_๐Ÿ โˆ’ ๐’›_๐Ÿ ) Given the line passes through the points So, the equation of line is (๐‘ฅ โˆ’ 3)/(5 โˆ’ 3) = (๐‘ฆ โˆ’ 4)/(1 โˆ’ 4) = (๐‘ง โˆ’ 1)/(6 โˆ’ 1) A (3, 4, 1) โˆด ๐‘ฅ_1= 3, ๐‘ฆ_1= 4, ๐‘ง_1= 1 B(5, 1, 6) โˆด ๐‘ฅ_2 = 5, ๐‘ฆ_2= 1, ๐‘ง_2= 6 (๐‘ฅ โˆ’ 3)/2 = (๐‘ฆ โˆ’ 4)/(โˆ’3) = (๐‘ง โˆ’ 1)/5 = k So, Since, the line crosses the XY plane at (x, y, 0) z = 0 5k + 1 = 0 5k = โˆ’1 โˆด k = (โˆ’๐Ÿ)/๐Ÿ“ So, x = 2k + 3 = 2 ร— (โˆ’1)/5 + 3 = 3 โˆ’ 2/5 = 13/5 y = โˆ’3 ร— (โˆ’1)/5 + 4 = 4 + 3/5 = 23/5 Therefore, the coordinates of the required point are (๐Ÿ๐Ÿ‘/๐Ÿ“, ๐Ÿ๐Ÿ‘/๐Ÿ“, ๐ŸŽ).

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.